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Math Help - doublte integral question

  1. #1
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    doublte integral question

    "The double integral \iint_{R}^{} \, f(x,y) dx\,dy can be expressed as \int_{\pi}^{2\pi} \int_{1}^{2} r^2 sin(\theta) dr\,d\Theta in polar coordinates. What is the region R? What is the formula for f(x,y)? What is the integral in cartesian coordinates?
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    Quote Originally Posted by tennisfreak412 View Post
    "The double integral \iint_{R}^{} \, dx\,dy can be expressed as \int_{Pi}^{2Pi} \int_{1}^{2} r^2 dr\,dTheta
    ummm...what's the question? are you supposed to find the region R?

    by the way, type \pi to get \pi and \theta to get \theta
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    sorry.. question is here:

    "The double integral \iint_{R}^{} \, f(x,y) dx\,dy can be expressed as \int_{\pi}^{2\pi} \int_{1}^{2} r^2 sin(\theta) dr\,d\Theta in polar coordinates. What is the region R? What is the formula for f(x,y)? What is the integral in cartesian coordinates?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tennisfreak412 View Post
    "The double integral \iint_{R}^{} \, f(x,y) dx\,dy can be expressed as \int_{\pi}^{2\pi} \int_{1}^{2} r^2 sin(\theta) dr\,d\Theta in polar coordinates. What is the region R? What is the formula for f(x,y)? What is the integral in cartesian coordinates?
    the radius goes from 1 to 2, the angle theta goes from pi to 2pi. thus the region R is the bottom half of the annulus formed by the circles with common center the origin, and radii 1 and 2 (see the diagram below).

    since \int \int f(x,y)~dxdy \to \int \int f(r \cos \theta, r \sin \theta)~rdrd \theta we see that the integral in polar coordinates is r. and we know that r = \sqrt{x^2 + y^2}

    i leave the rest to you

    can you continue?
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    I got that, but the difficulty I'm having is doing the double integral for the annulus. I know that I can have the dy limits going from -2 to 0, right? but then how do I do the dx limits so that they incorporate both sides of the hole in the annulus?
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    Quote Originally Posted by tennisfreak412 View Post
    I got that, but the difficulty I'm having is doing the double integral for the annulus. I know that I can have the dy limits going from -2 to 0, right? but then how do I do the dx limits so that they incorporate both sides of the hole in the annulus?
    Hint:

    the equation for the inner semi-circle is: - \sqrt{1 - x^2}

    the equation for the outer semi-circle is: - \sqrt{4 - x^2}

    can you continue?
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